Mean Value Theorem Solution 7

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SOLUTION 7: We are given the function

$f(x) = 2x + \sqrt{x-1}$ and the interval

$[1, 5]$ . This function is continuous on the closed interval

$[1, 5]$ since it is the sum of continuous functions

$y=2x$ (polynomial) and

$y= \sqrt{x-1}$

$\Big($ functional composition of continuous functions

$y=x-1$ (polynomial) and

$y= \sqrt{x}$ (well-known).

$\Big)$ . The derivative of

$f$ is

$f'(x) = 2 + (1/2)(x-1)^{-1/2} = 2 + \displaystyle{ 1 \over 2\sqrt{x-1} }$ We can now see that

$f$ is differentiable on the open interval

$(1, 5)$ . The assumptions of the Mean Value Theorem have now been met. Let's apply the Mean Value Theorem and find all possible values of

$c$ in the open interval

$(1, 5)$ . Then

$f'(c)= \displaystyle{ f(5)-f(1) \over 5 - 1 } \ \ \ \ \longrightarrow$

$2 + \displaystyle{ 1 \over 2 \sqrt{c-1} } = \displaystyle{ (2(5) + \sqrt{5-1} \ ) - (2(1) + \sqrt{1-1} \ ) \over 4 } \ \ \ \ \longrightarrow$

$2 + \displaystyle{ 1 \over 2 \sqrt{c-1} } = \displaystyle{ (10 + 2 ) - ( 2 + 0) \over 4 } \ \ \ \ \longrightarrow$

$2 + \displaystyle{ 1 \over 2 \sqrt{c-1} } = \displaystyle{ 5 \over 2 } \ \ \ \ \longrightarrow$

$\displaystyle{ 1 \over 2 \sqrt{c-1} } = \displaystyle{ 1 \over 2 } \ \ \ \ \longrightarrow$

$\sqrt{c-1} = 1 \ \ \ \ \longrightarrow$

$(\sqrt{c-1} \ )^2 = 1^2 \ \ \ \ \longrightarrow$

$c-1 = 1 \ \ \ \ \longrightarrow$

$c = 2$

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